What the is the polar form of $y = 1/y-xy+x^2/y$ ?

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Answer 1 · 2018-06-27T01:05:34

$rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/(sin(theta))$

Simplified: $r^2(cos(theta)-cos(3theta))-4=(2r)^2cos(2theta)$

$x=rcos(theta), y=rsin(theta)$

Substitute and Simplify

$rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/sin(theta)$

Alternate: $r^2sin(theta)=csc(theta)-r^3sin(theta)cos(theta)+r^2cot(theta)cos(theta)$

What the is the polar form of y = 1/y-xy+x^2/y y = 1/y-xy+x^2/y ?